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Question 123888: Ryan has 10-cent coins, 20-cent coins and 50-cent coins in his coin bank. 46 of them are not 20-cent coins, 54 of the coins are not 50-cent coins. The total number of 50-cent coins and 20-cent coins are 40. Find the total value of all the coins in his coin bank.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Ryan has 10-cent coins, 20-cent coins and 50-cent coins in his coin bank. 46 of them are not 20-cent coins, 54 of the coins are not 50-cent coins. The total number of 50-cent coins and 20-cent coins are 40. Find the total value of all the coins in his coin bank.
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Let x = 10-cent coins; Let y = 20-cent coins; Let z = 50-cent coins
Let t = total coins
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Ryan has 10-cent coins, 20-cent coins and 50-cent coins in his coin bank.
x + y + z = t
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46 of them are not 20-cent coins,
y = t - 46
y + 46 = t
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54 of the coins are not 50-cent coins.
z = t - 54
z + 54 = t
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The total number of 50-cent coins and 20-cent coins are 40.
y + z = 40
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Let's take the two equations that = t and make them = each other
y + 46 = z + 54
y - z = 54 - 46
y - z = 8
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pair this up with
y + z = 40
y - z = 8
-------------adding eliminates z
2y = 48
y = 24 ea 20-cent coins
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Find z using y + z = 40
24 + z = 40
z = 40 - 24
z = 16 ea 50-cent coins
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But what about x?
Substitute for y and z in the 1st equation
x + 24 + 16 = t
x + 40 = t
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We also know that z + 54 = t and z = 16; therefore t = 54 + 16 = 70
Using 70 for t:
x + 40 = 70
x = 70 - 40
x = 30 ea 10-cent coins
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We have x = 30; y = 24; z = 16
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Find the total value of all the coins in his coin bank.
10(30) + 20(24) + 50(16) =
300 + 480 + 800 = 1580 cents; $15.80
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